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10 Sep 2016, 10:54
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What is the area of the area of the triangle formed by the intersection of y = 2x-2, y = -1/2x + 8 and y=0 ?
a)45
b)67
c)89
d)31
e)37
a)45
b)67
c)89
d)31
e)37
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10 Sep 2016, 11:03
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PawanRamnani1234
What is the area of the area of the triangle formed by the intersection of y = 2x-2, y = -1/2x + 8 and y=0 ?
a)45
b)67
c)89
d)31
e)37
a)45
b)67
c)89
d)31
e)37
Answer A.
We are given the lines
y = 2x-2 --> X-intercept = 1 and y-intercept = -2
y = -1/2x + 8 --> X-intercept = 16 and y-intercept = 8
y=0 means x axis.
By drawing the lines on the X-Y plane, we will get the intersection of 1st two lines at (4,6).
So area = 1/2 * base * height = 1/2 * (16-1) * 6 = 45.
10 Sep 2016, 20:50
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abhimahna
PawanRamnani1234
What is the area of the area of the triangle formed by the intersection of y = 2x-2, y = -1/2x + 8 and y=0 ?
a)45
b)67
c)89
d)31
e)37
a)45
b)67
c)89
d)31
e)37
Answer A.
We are given the lines
y = 2x-2 --> X-intercept = 1 and y-intercept = -2
y = -1/2x + 8 --> X-intercept = 16 and y-intercept = 8
y=0 means x axis.
By drawing the lines on the X-Y plane, we will get the intersection of 1st two lines at (4,6).
So area = 1/2 * base * height = 1/2 * (16-1) * 6 = 45.
Hello,
Thanks for solution, I have two queries,
how did you concluded the intersection of first two lines at (4,6)?
why have you subracted 1 from 16?
